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Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car B at −10 m/s when they collide head–on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision? −8 m/s −10 m/s 0 m/s 10 m/s

Respuesta :

ANSWER:

-10 m/s

STEP-BY-STEP EXPLANATION:

We have the following information:

[tex]\begin{gathered} m_A=m_B=100\text{ kg} \\ v_{A1}=8\text{ m/s} \\ v_{B1}=-10\text{ m/s} \\ v_{B2}=8\text{ m/s} \end{gathered}[/tex]

Now by momentum conservation equation, we have:

[tex]\begin{gathered} m_A\cdot v_{A1}+m_2\cdot v_{B1}=m_A\cdot v_{A2}+m_2\cdot v_{B2} \\ \text{replacing and solving for }v_{A2}\colon \\ 100\cdot8+100\cdot-10=100\cdot v_{A2}+100\cdot8 \\ 8-10=v_{A2}+8 \\ v_{A2}=8-10-8 \\ v_{A2}=-10 \end{gathered}[/tex]

so car A will move with a velocity of -10 m/s

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